Question

tìm x:

\(\sqrt{x-2\sqrt{x-1}}=\sqrt{x-1}-1\)

\(\sqrt{9x^2-3x+\frac{1}{4}}=\frac{1}{2}-3x\)

THANKS FOR YOUR HELP ^-^

Answer 1

a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge1\\x\ge2\end{matrix}\right.\Leftrightarrow x\ge2\)

Ta có: \(\sqrt{x-2\sqrt{x-1}}=\sqrt{x-1}-1\)

\(\Leftrightarrow\sqrt{x-1-2\sqrt{x-1}\cdot1+1}-\sqrt{x-1}+1=0\)

\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-1\right)^2}-\sqrt{x-1}+1=0\)

\(\Leftrightarrow\left|\sqrt{x-1}-1\right|-\sqrt{x-1}+1=0\)

\(\Leftrightarrow\sqrt{x-1}-1-\sqrt{x-1}+1=0\)

\(\Leftrightarrow0x=0\)

Vậy: S={x|\(x\in R\)}